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Magnetic flux density - class-XII

Description: magnetic flux density
Number of Questions: 36
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Tags: physics electromagnetic induction electromagnetic induction and alternating currents magnetic fields and electromagnetism
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A square loop of side 12 cm and resistance 0.60$\Omega$ is placed vertically in the east-west plane. A uniform magnetic field of 0.1 T is setup across the plane in north-east direction. The magnetic field is decreased to zero in 0.6 s at a steady rate. The magnitude of current during this time interval is

  1. $1.42 \times 10^{-3} A$

  2. $2.67 \times 10^{-3} A$

  3. $3.41\times 10^{-3} A$

  4. $4.21 \times 10^{-3} A$


Correct Option: B
Explanation:

Here, Area $A=l^2=(12cm)^2=1.4\times 10^{-2} m^2$
$R=0.60\omega, B _1=0.10 T,\theta=45^0$
$B _2=0,dt=0.6$ s
Initial flux,
$\phi _1=B _1Acos\theta$
      $=0.10\times1.4\times10^{-2}\times cos 45^0$
      $=9.8\times 10^{-4}$

final flux, $\phi _2$=0
Induced emf,$E=\dfrac{| d\phi |}{dt}=\dfrac{|\phi _2-\phi _1|}{dt}$

                      $E=\dfrac{|9.8\times 10^{-4}|}{0.6}s\\,\,\,\,\,=1.6\times10^{-3}V$

Current, $I=\dfrac{E}{R}=\dfrac{1.6\times\times 10^{-3}}{0.6}=2.67 \times 10^{-3}$

When the normal to a coil points in the direction of magnetic field (B), then flux is 

  1. a scalar quantity

  2. a vector quantity

  3. neither scalar nor vector

  4. uncertain


Correct Option: A
Explanation:

Dot product of field and area vectors is flux . $\Phi=B.dS$, and we know dot product of two vectors is a scalar quantity.
Therefore, flux is scalar.

Current $i _0$ is being carried by an infinite wire passing through origin along the direction $\hat{i} + \hat{j} + \hat{k}$. Find magnetic field due to the wire at point $(1 m, 0, 0)$.

  1. $\dfrac{(\mu _0 i)}{(2 \pi)} T$

  2. $\dfrac{(\mu _0 i)}{(\sqrt{2} \pi)} T$

  3. $\dfrac{(\mu _0 i)}{(4 \pi)} T$

  4. $\dfrac{(\sqrt{3} \mu _0 i)}{(2 \sqrt{2} \pi)} T$


Correct Option: D

A charge q is placed at the centre of a cylinder of radius R and length 2R. Then electric flux through the curved surface of the cylinder is 

  1. $\cfrac { q }{ 2 { \epsilon } _{ 0 } } $

  2. $\cfrac { q }{ 4 { \epsilon } _{ 0 } } $

  3. $\cfrac { q }{ \sqrt { 2 } { \epsilon } _{ 0 } } $

  4. $\cfrac { q }{ 2\sqrt { 2 } { \epsilon } _{ 0 } } $


Correct Option: C

A Point source generates 10 J of light energy in 2 s. The luminous flux of source is: 

  1. 5 lumen

  2. 10 lumen

  3. 50 lumen

  4. none of these


Correct Option: A
Explanation:

The luminous flux of source is the rate at which light energy flows from the source:

luminous flux =10 J/ 2 s= 5 lumen.
so the correct option is A.

A cyclotron in which protons are accelerated has a flux density 1.57T. The variation of frequency of electric field is (in Hz) 

  1. $4.8 \times 10 ^ { 8 }$

  2. $8.4 \times 10 ^ { 8 }$

  3. $2.5 \times 10 ^ { 7 }$

  4. $4.8 \times 10 ^ { 6 }$


Correct Option: C

The electric field potential in space has the form $V(x,y,z)=-2xy+3yz^{-1}$. The electric field intensity $\vec E$ magnitude at the point (-1,1,2) is

  1. $2 \sqrt{86} units$

  2. $2\sqrt{163} units$

  3. $\sqrt{163} units$

  4. $ \sqrt{86} units$


Correct Option: A

An iron rod of volume ${ 10 }^{ -4 }{ m }^{ 3 }$ and relative permeability 1000 is placed inside a long solenoid would with. 5 turn ism. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod is:

  1. 10 ${ Am }^{ 2 }$

  2. 15 ${ Am }^{ 2 }$

  3. 20 ${ Am }^{ 2 }$

  4. 25 ${ Am }^{ 2 }$


Correct Option: D

Unit of magnetic flux density is 

  1. $weber/metre$

  2. $weber$

  3. $weber/m^{2}$

  4. $ampere/m$


Correct Option: C

Current in a circular coil having negligible resistance and inductance 0.1 H is increasing at the rate of $1 As^{-1}$. The power generated in the coil when the magnetic flux linked with it is 0.1 Wb will be:-

  1. 0.05 W

  2. 0.1W

  3. 1 W

  4. 10 W


Correct Option: B

The magnetic flux density at a point distant $d$ from a long straight current carrying conductor is $B$, then its value at distance $d/2$ will be:

  1. $4B$

  2. $2B$

  3. $B/2$

  4. $B/4$


Correct Option: B
Explanation:

Flux density, B $=\dfrac{\phi}{A}$
$B _A=\dfrac{\mu _0 I}{2\pi d}=B$
$=\dfrac{\phi}{A}$
$B _A=\dfrac{\mu _0 I}{2\pi \dfrac{d}{2}}=2\dfrac{\mu _0 I}{2\pi d}$
$B _B=2\times B$
$B _B=2B$

The magnetic needle of a tangent galvanometer is deflected at an angle $30$ due to a magnet. The horizontal component of earth's magnetic field $0.34\times 10^{-4}T$ is along the plane of the coil. The magnetic intensity is:

  1. $1.96\times 10^{-4}T$

  2. $1.96\times 10^{-5}T$

  3. $1.96\times 10^{4}T$

  4. $1.96\times 10^{5}T$


Correct Option: B
Explanation:

The correct option is B.


Given,

$B=0.34\times10^{-4}T$

Deflected angle$\theta=30^0$

So magnetic intensity is $Btan 30^0$

$=0.34\times10^{-4}T\times\dfrac{1}{\sqrt3}$

$=1.96\times10^{-5}T$

Where,$tan 30^0=\dfrac{1}{\sqrt3}$, and $\sqrt3=1.73$c

A sphere of radius $R$ and charge $Q$ is placed inside an imaginary sphere of radius $2R$. Whose center coincides with the given sphere. The flux related to the imaginary sphere is:

  1. $\dfrac {Q}{\in _{0}}$

  2. $\dfrac {Q}{2\in _{0}}$

  3. $\dfrac {4Q}{\in _{0}}$

  4. $\dfrac {2Q}{\in _{0}}$


Correct Option: A

The flux linked with a coil changes with time according to the equation $\phi$ = a$t^2$ +bt +c. Then SI unit of a is 

  1. Volt

  2. Volt/sec

  3. Volt.sec

  4. Weber


Correct Option: B
Explanation:

The unit of $\phi$ is Volt-sec.

Now $\phi=a t^{2}+bt+c$
To meet the dimension requirement $at^{2}$ must be Volt-sec.
$a$ must Volt/sec.

In a circuit a coil of resistance $2\,\Omega$, then magnetic flux charges from $2.0\,Wb$ to $10.0\,Wb$ in $0.2\ sec.$ The charge flow in the coil during this time is:

  1. $5.0\ C$

  2. $4.0\ C$

  3. $1.0\ C$

  4. $0.8\ C$


Correct Option: B
Explanation:

The relation between the rate of change of charge (or current) and the flux is given by the following relation:  

$ \because \dfrac{dQ}{dt}=-\dfrac{1}{R}\dfrac{d\phi }{dt} $

$ \dfrac{dQ}{dt}=\dfrac{-(10-2)}{2}=4\,C $


Two coils $A$ and $B$ are wound on the same iron  core as shown in figure. The number of turns in the coil $A$ and $B$ are $N _{A}$ and $N _{B}$ respectively. Identity the correct statement 

  1. Both the coils have same magnitude of magnetic flux

  2. The magnetic flux linked are in the ratio $\dfrac{\phi A}{\phi B}=\dfrac{N _{A}}{N _{B}}$

  3. The induced emf across each coil are in the ratio $\dfrac{E _{A}}{E _{B}}=\left(\dfrac{N _{4}}{N _{B}}\right)^{2}$

  4. Both the coils have same magnitude of induced emf


Correct Option: B

The magnetic flux through a stationary loop with resistance R varies during the interval of time T as $\phi  = at(T - t)$ ./ The heat generated during this time neglecting the inductance of the loop will be :

  1. $\dfrac{{{a^2}{T^3}}}{{3R}}$

  2. $\dfrac{{{a^2}{T^2}}}{{3R}}$

  3. $\dfrac{{{a^2}T}}{{3R}}$

  4. $\dfrac{{{a^3}{T^3}}}{{3R}}$


Correct Option: C

A circular disc of radius $0.2$m isplaced in a uniform magnetic field of induction $\dfrac{1}{\pi}\left(\dfrac {Wb}{m^2}\right)$ in such a way that its axis ,makes an angle of $60^o$ with $\xrightarrow {B}$. The magnetic flux linked with the disc is

  1. $0.01$ Wb

  2. $0.02$ Wb

  3. $0.06$ Wb

  4. $0.08$ Wb


Correct Option: A

The magnetic flux through a coil is $4\times 10^{-4} W/b/m^2$ at time $t=0$.It reduces to $10\%$ of its original value in 't' seconds.If the induced e.m.f is $0.72 m V,$ then the time t is:

  1. $0.25 s$

  2. $0.05 s$

  3. $0.75 s$

  4. $1 s$


Correct Option: B
Explanation:

As given in question

magnetic turn $=(\phi _1) = 4 \times 10^{-4}wb/m^2$
at $t = 0$
At, $t =t _2$, the turn reduces to $10\%$, means 
$\phi _2 = 0.9\ \phi _1$
As, per the farady's law,
In duced Emf $= \dfrac{Nd\ \phi}{dt}$
$e = \dfrac{nd\phi}{dt}$      ...(1)
$e = 0.72mu$ pur in (1), take $N = 1$ turns are constant
$0.72 \times 10^{-3} = \dfrac{-(\phi _2-\phi _1)}{(t _2-t _1)}$
$0.72\times 10^{-3} = \dfrac{-(0.9-2)\phi _1}{(t _2-0)}$
$t _2 = \dfrac{(0.1)\times (4\times 10^{-4})}{(0.72\times 10^{-3})}$
$t _2 = 0.05\ sec$

In a uniform electric field $\vec {E}$ an imaginary cube of edge length $a$ is considered as shown. The outward flux linked with cube surface will be :

  1. $Ea^{2}$

  2. $\sqrt{2}Ea^{2}$

  3. $\sqrt{3}Ea^{2}$

  4. $2Ea^{2}$


Correct Option: A

State whether the following two statements are true or false
(i) Li has the same units as that of magnetic flux.
(ii) Li has the units volt-second and magnetic flux has the units coulomb-ohm.

  1. T T

  2. F F

  3. T F

  4. F T


Correct Option: C
Explanation:

$\begin{array}{l} \left( 1 \right) This\, \, is\, \, True\, \, because\, \, \phi =Li \ \left( 2 \right) This\, \, is\, \, False\, \, because \ v=L\left( { \dfrac { { di } }{ { dt } }  } \right)  \ \Rightarrow L=\dfrac { v }{ q }  \ \Rightarrow Li=\dfrac { v }{ B }  \ Hence, \ option\, \, C\, \, is\, correct\, \, answer. \end{array}$

The ratio of magnetic inductions at the centre of a circular coil of radius a and on its axis at a distance equal to its radius, will be -

  1. $\frac { 1 }{ \sqrt { 2 } } $

  2. $\frac { \sqrt { 2 } }{ 1 } $

  3. $\frac { 1 }{ 2\sqrt { 2 } } $

  4. $\frac { 2\sqrt { 2 } }{ 1 } $


Correct Option: C

The sun delivers ${10^3}W/{m^2}$ of electromagnetic flux to the earth's surface. The solar energy incident on the roof in $1$houre will be 

  1. $5.76 \times {10^8}J$

  2. $5.76 \times {10^7}J$

  3. $5.76 \times {10^6}J$

  4. $5.76 \times {10^5}J$


Correct Option: C

Light with an energy flux of $18 w/cm^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20 cm^2$. Find the average force exerted on the surface during a 30 minute time

  1. $1.2 \times 10 ^ { - 6 } N$

  2. $2 .4\times 10 ^ { - 6 } N$

  3. $2.16 \times 10 ^ { - 3 } \mathrm { N }$

  4. $1.5\times 10 ^ { - 6 } N$


Correct Option: A
Explanation:

The total energy falling on the surface is $U = \left( {18W/c{m^2}} \right) \times \left( {20c{m^2}} \right) \times \left( {30 \times 60} \right) = 6.48 \times {10^5}J$

therefore$,$ the total momentum delivered is 
$P = \dfrac{U}{c} = \dfrac{{\left( {6.48 \times {{10}^5}J} \right)}}{{\left( {3 \times {{10}^8}\,m/s} \right)}} = 2.16 \times {10^{ - 3}}\,kgm/s$
The average force exerted on the surface is 
$F = \dfrac{p}{t} =  = \dfrac{{2.16 \times {{10}^{ - 3}}}}{{0.18 \times {{10}^4}}} = 1.2 \times {10^{ - 6}}N$
Hence,
option $(A)$ is correct answer.

The electric field in a certain region is $\left( 10\hat { i } +5\hat { j }  \right) \times { 10 }^{ 4 }N/C$. What is the flux due to this field over an area of $\left( 3\hat { i } +3\hat { j }  \right) \times { 10 }^{ -2 }{ m }^{ 2 }$ in ${ Nm }^{ 2 }/C?$

  1. $4.5\times { 10 }^{ 3 }$

  2. $3.5\times { 10 }^{ 3 }$

  3. $2.5\times { 10 }^{ 3 }$

  4. $1.5\times { 10 }^{ 3 }$


Correct Option: A

Current flowing through a long solenoid is varied. Then, magnetic flux density of the magnetic field inside varies ::

  1. inversely with $I$

  2. inversely with ${ I }^{ 2 }$

  3. directly with $I$

  4. directly with ${ I }^{ 2 }$


Correct Option: C

A wire of length $'\ell '$ is used to make a circular coil of 'n' no. of turns. A current 'I' passes through the coil. If twice the length of wire is used now to make the coil with turns of same radius, making same current flow through it, the magnetic field at the centre of coil will become.

  1. 1.5 times than earlier

  2. 0.5 time than earlier

  3. 2 times than earlier

  4. 4 time than earlier


Correct Option: B

The magnetic flux linked with a coil is given by the equation $ \phi = 5t^2 + 3t + 6 $. The induced e.m.f. in the coil in the fourth second will be 

  1. 20 V

  2. 40 V

  3. 10 V

  4. 80 V


Correct Option: C

A closely wound flat circular coil of 25 turns of wire has diameter of =f 10 cm and carries a current of 4 amperes. determine the magnetic flux density at the centre of the coil:-

  1. $ 1.679 \times 10^{-5} T $

  2. $ 2.028 \times 10^{-4} T $

  3. $ 1.257 \times 10^{-3} T $

  4. $ 1.512 \times 10^{-6} T $


Correct Option: C

The magnetic flux linked with a coil, in webers, is given by the equation $\phi =4t^{2}-3t+7$. Then the magnitude of induced emf at 2 sec will be....

  1. 15 v

  2. 19 v

  3. 17 v

  4. 21 v


Correct Option: C

Magnetic field intensity at the centre of coil of 50 turns, radius 0.5 m and carrying a current of 2 A is 

  1. $0.5 \times 10^{-5}T$

  2. $3 \times 10^{-5}T$

  3. $1.25 \times 10^{-4}T$

  4. $4 \times 10^{-5}T$


Correct Option: C

Select the incorrect option

  1. Luminous flux and radiant flux have same dimensions

  2. Luminous flux and luminous intensity have same dimensions

  3. Radiant flux and power have same dimension

  4. Relative luminosity is a dimensionless quantity


Correct Option: A
Explanation:

Luminous flux ( Luminous power) is the amount of perceived power of light. It is measure energy.

Radiant flux (Radiant power) is the amount of radiated power.
While luminous intensity is ratio of luminous flux to solid angle.

A current carrying wire produces a magnetic field in its surrounding space.
The S.I. unit of magnetic flux density is 

  1. Henry

  2. Tesla

  3. $AM^2$

  4. A-m


Correct Option: B
Explanation:

The SI unit of Magnetic Flux density is $ Weber/m^2 $ which is also called Tesla.

The dimensional formula of magnetic flux is ___________.

  1. $\left[ M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -1 } \right] $

  2. $\left[ M{ L }^{ 2 }{ T }^{ -3 }{ A }^{ -1 } \right] $

  3. $\left[ { M }^{ -1 }{ L }^{ -2 }{ T }^{ 2 }{ A }^{ 1 } \right] $

  4. $\left[ M{ L }^{ 3 }{ T }^{ -2 }{ A }^{ -1 } \right] $


Correct Option: A
Explanation:

Dimensions of ${ \phi  } _{ B }=$ Unit of $B\times $ Unit of $A$
$=\dfrac { newton }{ ampere-metre } \times { metre }^{ 2 }=\dfrac { newton\times metre }{ ampere } $
$=\dfrac { kg{ ms }^{ -2 }\times m }{ A } =kg{ m }^{ 2 }{ s }^{ -2 }{ A }^{ -1 }=\left[ M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -1 } \right] $

In an experiment to measure the velocity of electrons, an electric field $(E)$ and a magnetic field $(B)$ are employed to produce zero deflection. Then :

  1. The two fields are parallel and the velocity is given by $BE$

  2. The two fields are perpendicular and the velocity is given by $E/B$

  3. The two fields are parallel and the velocity is given by $E/B$

  4. The two fields are perpendicular and the velocity is given by $BE$


Correct Option: B
Explanation:

The electric force on electron of charge ${e}$  is,
             $  F= {e}E $ 
                    where, E = electric field.

And Magnetic force ,
            $ F= {e}$ $(V\times B)$
               where, V= velocity of electron in magnetic field
                            B= magnetic field

From the above equations,

          $E= V\times B$
Electric field E is perpendicular to both velocity of electron and the magnetic field. So,  The two field is perpendicular to each other and the velocity of electron is $\dfrac{E}{B}$.

B. The two fields are perpendicular and the velocity is given by $\dfrac{E}{B}$.

Write the dimensions of Magnetic flux in terms of mass, time, length and charge.

  1. $[M^1L^2T^{-1}Q^{-2}]$

  2. $[M^1L^2T^{-1}Q^{-1}]$

  3. $[M^1L^3T^{-1}Q^{-1}]$

  4. $[M^4L^2T^{-1}Q^{-1}]$


Correct Option: B
Explanation:

B. $[M^1 L^2 T^ {-1}Q^{-1}]$


Dimension of magnetic field, 


$[B]=[M^1L^0Q^{-1}]$ 

Dimension of surface Area,

 $[A]=[L^2]$

The magnetic flux , 

$\phi=B.A$

The dimension of magnetic flux in term of mass, time, length and charge,
$[\phi]= [M^1L^2T^{-1}Q^{-1}]$

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